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Chapter 8: Problem 47

In Problems \(43-50\), use an inverse matrix to solve the given system of equations. $$ \begin{aligned} x_{1}+x_{3} &=-4 \\ x_{1}+x_{2}+x_{3} &=0 \\ 5 x_{1}-x_{2} &=6 \end{aligned} $$

Short Answer

Expert verified
The solution is \( x_1 = -10 \), \( x_2 = -26 \), \( x_3 = 10 \).

Step by step solution

01

Write the System in Matrix Form

First, express the system of equations as a matrix equation. The given system can be written as: \[ A \mathbf{x} = \mathbf{b} \] where \[ A = \begin{bmatrix} 1 & 0 & 1 \ 1 & 1 & 1 \ 5 & -1 & 0 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}, \quad \text{and} \quad \mathbf{b} = \begin{bmatrix} -4 \ 0 \ 6 \end{bmatrix} \].
02

Find the Inverse of Matrix A

Calculate the inverse of the matrix \( A \), denoted as \( A^{-1} \). The inverse exists only if \( A \) is a square matrix (which it is) and has a non-zero determinant. After calculation, we find: \[ A^{-1} = \begin{bmatrix} 1 & -4 & -1 \ 5 & -9 & -1 \ -1 & 5 & 1 \end{bmatrix} \].
03

Solve for \( \mathbf{x} \) Using the Inverse Matrix

Multiply both sides of the matrix equation by \( A^{-1} \) to solve for \( \mathbf{x} \): \[ A^{-1} A \mathbf{x} = A^{-1} \mathbf{b} \] which simplifies to \[ \mathbf{x} = A^{-1} \mathbf{b} \]. Substituting the known matrices gives us: \[ \mathbf{x} = \begin{bmatrix} 1 & -4 & -1 \ 5 & -9 & -1 \ -1 & 5 & 1 \end{bmatrix} \begin{bmatrix} -4 \ 0 \ 6 \end{bmatrix} \].
04

Perform Matrix Multiplication

Calculate the resulting vector from the multiplication: \[ \begin{bmatrix} 1 & -4 & -1 \ 5 & -9 & -1 \ -1 & 5 & 1 \end{bmatrix} \begin{bmatrix} -4 \ 0 \ 6 \end{bmatrix} = \begin{bmatrix} 1(-4) + (-4)(0) + (-1)(6) \ 5(-4) + (-9)(0) + (-1)(6) \ -1(-4) + 5(0) + 1(6) \end{bmatrix} = \begin{bmatrix} -4 - 6 \ -20 - 6 \ 4 + 6 \end{bmatrix} = \begin{bmatrix} -10 \ -26 \ 10 \end{bmatrix} \].
05

Interpret the Resulting Vector

The vector \( \begin{bmatrix} -10 \ -26 \ 10 \end{bmatrix} \) is the solution to the system of equations. This means \( x_1 = -10 \), \( x_2 = -26 \), and \( x_3 = 10 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication
Matrix multiplication is a method used to calculate a new matrix from two matrices. It's a bit different from how we multiply regular numbers. Here's the process to understand it better:
  • To multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix.
  • Multiplication involves taking each row in the first matrix and multiplying it by each column in the second matrix.
  • The result is a new matrix, where each element is a sum of the products of corresponding elements from the row and column.
For example, in our exercise, we multiply matrix \( A^{-1} = \begin{bmatrix} 1 & -4 & -1 \ 5 & -9 & -1 \ -1 & 5 & 1 \end{bmatrix} \) by matrix \( \mathbf{b} = \begin{bmatrix} -4 \ 0 \ 6 \end{bmatrix} \). We take each row from \( A^{-1} \) and carry out the dot product with \( \mathbf{b} \). That's like calculating \( 1(-4) + (-4)(0) + (-1)(6) \) for the first element of the result. This process is repeated for each row.
Determinant of a Matrix
The determinant is a special number that can be calculated from a square matrix. It's a key part of finding the inverse of a matrix.
  • The determinant helps in understanding if a matrix has an inverse. If it's zero, the matrix doesn't have an inverse.
  • The calculation involves subtracting the products of diagonal elements in larger matrices, and it's a bit complex for bigger sizes.
For our triangle-shaped matrix \( A \), calculating the determinant verifies that our matrix is invertible. This ensures that we can safely proceed to find \( A^{-1} \) and solve the system.
System of Equations
A system of equations consists of multiple equations with multiple variables. The goal is to find the values of these variables that satisfy all equations simultaneously.
  • Each equation can be thought of as a straight line (or plane in higher dimensions) and we look for the intersection points (solutions).
  • Solving using matrices helps to visualize and simplify complex systems into a structured manner.
In our exercise, the system has three equations with three unknowns: \( x_1, x_2, \) and \( x_3 \). Transforming it into the matrix form \( A\mathbf{x} = \mathbf{b} \) allows leveraging linear algebra methods like inverse matrices for solutions.
Matrix Equation
A matrix equation is a compact way to represent systems of linear equations. It uses matrices to encode the coefficients from the equations, which simplifies calculations and solution techniques.
  • The typical form is \( A\mathbf{x} = \mathbf{b} \), where \( A \) is a matrix of coefficients, \( \mathbf{x} \) is a vector of variables, and \( \mathbf{b} \) is a vector of equation results.
  • Finding solutions involves matrix operations, such as finding matrix inverses and performing matrix multiplications.
In our step-by-step solution, a matrix equation bridges the original system and its resolution, simplifying the process by offering a structured path to calculate \( \mathbf{x} \) using the inverse of \( A \).

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Most popular questions from this chapter

In Problems, find the eigenvalues and eigenvectors of the given nonsingular matrix \(\mathbf{A} .\) Then without finding \(\mathbf{A}^{-1}\), find its eigenvalues and corresponding eigenvectors. $$ A=\left(\begin{array}{rr} 4 & 2 \\ 7 & -1 \end{array}\right) $$

$$ \text { Write }\left(\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} b_{1} \\ b_{2} \end{array}\right) \text { without matrices. } $$

Suppose \(\lambda\) is an eigenvalue with corresponding eigenvector \(\mathbf{K}\) of an \(n \times n\) matrix \(\mathbf{A}\). (a) If \(\mathbf{A}^{2}=\mathbf{A} \mathbf{A}\), then show that \(\mathbf{A}^{2} \mathbf{K}=\lambda^{2} \mathbf{K}\). Explain the meaning of the last equation. (b) Verify the result obtained in part (a) for the matrix \(\mathbf{A}=\left(\begin{array}{ll}2 & 3 \\ 5 & 4\end{array}\right)\)

Use the inverse power method to find the eigenvalue of least magnitude for the given matrix. $$ \left(\begin{array}{ll} 1 & 1 \\ 3 & 4 \end{array}\right) $$

$$ \text { Show that if } \mathbf{A} \text { is an } m \times n \text { matrix, then } \mathbf{A A}^{T} \text { is symmetric. } $$
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